Question

A rocket accelerates straight up from the ground at 12.6 m/s^2 for 11.0 s. Then the engine cuts off and the rocket enters free fall. (a) Find its velocity at the end of its upward acceleration. (b) What maximum height does it reach? (c) With what velocity does it crash to Earth? (d) What's the total time from launch to crash?

Answer 1

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s$

Velocity at the end of its upward acceleration is 138.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m$

Maximum height the rocket reaches is 762.3 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s$

The velocity with which the rocket crashes to the Earth is 122.3 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s$

Total time from launch to crash is 12.47+11 = 24.47 seconds